How to integrate the dilogarithms?

Question

$\def\Li{\operatorname{Li}}$ How can you integrate $\Li_2$? I tried from $0 \to 1$

$\displaystyle \int_{0}^{1} \Li_2(z) \,dz = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)}$

$$\frac{An + B}{n^2} + \frac{D}{n+1} = \frac{1}{n^2(n+1)}$$

$$(An + B)(n+1) + D(n^2) = 1$$

Let $n = -1, \implies D = 1$ Let $n = 0, \implies B = 1$ Let $n = 1, \implies A = -1$

$$\frac{-n + 1}{n^2} + \frac{1}{n+1} = \frac{1}{n^2(n+1)}$$

$$= \sum_{n=1}^{\infty} \frac{-n + 1}{n^2} + \frac{1}{n+1} = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)} = \sum_{n=1}^{\infty} \frac{1}{n^2} - \frac{1}{n} + \frac{1}{n+1} $$

The $1/n$ is the problem, it is the harmonic series, which diverges.

Answer 1

$$ \begin{align} \int_0^1\mathrm{Li}_2(x)\,\mathrm{d}x &=\int_0^1\sum_{k=1}^\infty\frac{x^k}{k^2}\,\mathrm{d}x\\ &=\sum_{k=1}^\infty\frac1{(k+1)k^2}\\ &=\sum_{k=1}^\infty\left(\frac1{k^2}-\frac1k+\frac1{k+1}\right)\\ &=\zeta(2)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)\\ &=\frac{\pi^2}{6}-1 \end{align} $$ where $$ \begin{align} \sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right) &=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\sum_{k=2}^{n+1}\frac1k\right)\\ &=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)\\[12pt] &=1 \end{align} $$ is a telescoping series.

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Note that we only break $\left(\dfrac1k-\dfrac1{k+1}\right)$ into $\dfrac1k$ and $\dfrac1{k+1}$ when we are looking at finite sums. When considering infinite sums, it is kept together, and is equal to $\dfrac1{k(k+1)}$.

Answer 2

Maybe you should look at your decomposition as

$$\frac1{n^2 (n+1)} = \frac1{n^2} - \frac1{n (n+1)}$$

The sum over the second term is easy, given that the indefinite sum is telescoping, i.e.,

$$\sum_{n=1}^N \frac1{n (n+1)} = 1-\frac1{N+1}$$

We take the limit as $N \to \infty$ and then we may view this as the infinite sum. (Otherwise, as you say, there are convergence issues.)

Thus the sum in question is $$\frac{\pi^2}{6}-1$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large\int_{0}^{1}{\rm Li}_{2}\pars{x}\,\dd x} =\left.\vphantom{\Large A}x\,{\rm Li}_{2}\pars{x}\right\vert_{0}^{1} -\int_{0}^{1}x\,{\rm Li}_{2}'\pars{x}\,\dd x \\[5mm]&=\overbrace{{\rm Li}_{2}\pars{1}} ^{\ds{\color{#c00000}{\sum_{n\ =\ 1}^{\infty}{1^{n} \over n^{2}}\ =\ {\pi^{2} \over 6}}}}\ -\ \int_{0}^{1}x\ \overbrace{\bracks{-\,{{\ln\pars{1 - x} \over x}}}} ^{\ds{=\ \color{#c00000}{{\rm Li}_{2}'\pars{x}}}}\ \,\dd x ={\pi^{2} \over 6} +\ \underbrace{\int_{0}^{1}\ln\pars{x}\,\dd x}_{\ds{=\ \color{#c00000}{-1}}} \\[5mm]&=\color{#66f}{\large{\pi^{2} \over 6} - 1} \approx {\tt 0.6449} \end{align}